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, Chapter P
Fundamental Concepts of Algebra
Section P.1 6. a. 1− 2
Check Point Exercises Because 2 ≈ 1.4, the number inside the
absolute value bars is negative. The absolute
1. 8 + 6( x − 3)2 = 8 + 6(13 − 3)2 value of x when x < 0 is –x. Thus,
= 8 + 6(10) 2 (
1 − 2 = − 1 − 2 = 2 −1 )
= 8 + 6(100)
= 8 + 600 b. π −3
TU
= 608 Because π ≈ 3.14, the number inside the
absolute value bars is positive. The absolute
2. a. Since 2016 is 16 years after 2000, substitute 16 value of a positive number is the number itself.
for x. Thus,
T = − x 2 + 361x + 3193 π − 3 = π − 3.
= −(16) 2 + 361(16) + 3193
= 8713 x
TO
c.
The average cost of tuition and fees at public x
U.S. colleges for the school year ending in Because x > 0, x = x.
2016 was $8713.
x x
b. The formula underestimates the actual answer Thus, = =1
x x
by $65.
7. −4 − (5) = −9 = 9
3. The elements common to {3, 4, 5, 6, 7} and
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{3, 7, 8, 9} are 3 and 7. The distance between –4 and 5 is 9.
{3, 4,5, 6, 7} ∩ {3, 7,8,9} = {3, 7}
8. 7(4 x 2 + 3x) + 2(5 x 2 + x)
4. The union is the set containing all the elements of = 7(4 x 2 + 3 x) + 2(5 x 2 + x)
G
either set.
{3, 4,5, 6, 7} ∪ {3, 7,8,9} = {3, 4,5, 6, 7,8,9} = 28 x 2 + 21x + 10 x 2 + 2 x
= 38 x 2 + 23 x
π ½
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5. ® −9, − 1.3, 0, 0.3, , 9, 10 ¾
9. 6 + 4[7 − ( x − 2)]
¯ 2 ¿
= 6 + 4[7 − x + 2)]
a. Natural numbers: 9 because 9 =3 = 6 + 4[9 − x]
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= 6 + 36 − 4 x
b. Whole numbers: 0, 9 = 42 − 4 x
c. Integers: −9, 0, 9
U
Concept and Vocabulary Check P.1
d. Rational numbers: −9, − 1.3, 0, 0.3, 9
C1. expression
π C2. b to the nth power; base; exponent
e. Irrational numbers: , 10
2
C3. formula; modeling; models
π
f. Real numbers: −9, − 1.3, 0, 0.3, , 9, 10 C4. intersection; A ∩ B
2
C5. union; A ∪ B
Copyright © 2022 Pearson Education, Inc 1
R
U
G
R
TO
TU
, U
R
U
G
R
TO
TU
,TU
TO
The page is intentionally left blank
R
G
U
R
U
, Chapter P
Fundamental Concepts of Algebra
Section P.1 6. a. 1− 2
Check Point Exercises Because 2 ≈ 1.4, the number inside the
absolute value bars is negative. The absolute
1. 8 + 6( x − 3)2 = 8 + 6(13 − 3)2 value of x when x < 0 is –x. Thus,
= 8 + 6(10) 2 (
1 − 2 = − 1 − 2 = 2 −1 )
= 8 + 6(100)
= 8 + 600 b. π −3
TU
= 608 Because π ≈ 3.14, the number inside the
absolute value bars is positive. The absolute
2. a. Since 2016 is 16 years after 2000, substitute 16 value of a positive number is the number itself.
for x. Thus,
T = − x 2 + 361x + 3193 π − 3 = π − 3.
= −(16) 2 + 361(16) + 3193
= 8713 x
TO
c.
The average cost of tuition and fees at public x
U.S. colleges for the school year ending in Because x > 0, x = x.
2016 was $8713.
x x
b. The formula underestimates the actual answer Thus, = =1
x x
by $65.
7. −4 − (5) = −9 = 9
3. The elements common to {3, 4, 5, 6, 7} and
R
{3, 7, 8, 9} are 3 and 7. The distance between –4 and 5 is 9.
{3, 4,5, 6, 7} ∩ {3, 7,8,9} = {3, 7}
8. 7(4 x 2 + 3x) + 2(5 x 2 + x)
4. The union is the set containing all the elements of = 7(4 x 2 + 3 x) + 2(5 x 2 + x)
G
either set.
{3, 4,5, 6, 7} ∪ {3, 7,8,9} = {3, 4,5, 6, 7,8,9} = 28 x 2 + 21x + 10 x 2 + 2 x
= 38 x 2 + 23 x
π ½
U
5. ® −9, − 1.3, 0, 0.3, , 9, 10 ¾
9. 6 + 4[7 − ( x − 2)]
¯ 2 ¿
= 6 + 4[7 − x + 2)]
a. Natural numbers: 9 because 9 =3 = 6 + 4[9 − x]
R
= 6 + 36 − 4 x
b. Whole numbers: 0, 9 = 42 − 4 x
c. Integers: −9, 0, 9
U
Concept and Vocabulary Check P.1
d. Rational numbers: −9, − 1.3, 0, 0.3, 9
C1. expression
π C2. b to the nth power; base; exponent
e. Irrational numbers: , 10
2
C3. formula; modeling; models
π
f. Real numbers: −9, − 1.3, 0, 0.3, , 9, 10 C4. intersection; A ∩ B
2
C5. union; A ∪ B
Copyright © 2022 Pearson Education, Inc 1
